Find $\lim_{x\to 1}\dfrac{\sqrt{5x+4}-3}{x-1}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{3}{5}$ (Choice B) B $\dfrac{5}{6}$ (Choice C) C $1$ (Choice D) D The limit doesn't exist
Explanation: Substituting $x=1$ into $\dfrac{\sqrt{5x+4}-3}{x-1}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{\sqrt{5x+4}-3}{x-1} \\\\ &=\dfrac{\sqrt{5x+4}-3}{x-1}\cdot\dfrac{\sqrt{5x+4}+3}{\sqrt{5x+4}+3} \gray{\text{Rationalize the numerator}} \\\\ &=\dfrac{(5x+4)-3^2}{(x-1)(\sqrt{5x+4}+3)} \\\\ &=\dfrac{5\cancel{(x-1)}}{\cancel{(x-1)}(\sqrt{5x+4}+3)} \gray{\text{Cancel out common factors}} \\\\ &=\dfrac{5}{\sqrt{5x+4}+3} \text{, for }x\neq 1 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $1$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{\sqrt{5x+4}-3}{x-1}=\dfrac{5}{\sqrt{5x+4}+3}$ for all $x$ -values in the interval $(0.5,1.5)$ except for $x=1$. Therefore, $\lim_{x\to 1}\dfrac{\sqrt{5x+4}-3}{x-1}=\lim_{x\to 1}\dfrac{5}{\sqrt{5x+4}+3}=\dfrac{5}{6}$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to 1}\dfrac{\sqrt{5x+4}-3}{x-1}=\dfrac{5}{6}$.